Problem 21
Question
For the following exercises, graph the given functions by hand. $$ y=-|x|-2 $$
Step-by-Step Solution
Verified Answer
Graph a downward V-shape with vertex at (0,-2).
1Step 1: Understand the Basic Function Shape
The function given is based on the absolute value function, which has a characteristic V-shape. Here, you should understand that \(y = |x|\) is the base function. The absolute value function is symmetric around the y-axis and has a vertex at the origin (0,0) with two linear parts: \(y = x\) for \(x \geq 0\) and \(y = -x\) for \(x < 0\).
2Step 2: Consider Transformations
The given function \(y = -|x| - 2\) involves two transformations. First, the negative sign outside the absolute value flips the graph upside down (reflection over the x-axis). Then, the \(-2\) shifts the entire graph downward by 2 units. This means the new vertex of the graph is at (0,-2).
3Step 3: Plot the Vertex
After performing the transformations, plot the new vertex on the graph. The vertex is now at the point (0,-2). Start by plotting this point on the coordinate plane.
4Step 4: Draw the V-Shape
Using the transformations, draw the characteristic V-shape that opens downwards. From the vertex (0,-2), since the graph is flipped, draw two lines: one line with slope \(-1\) to the left and one line with slope \(1\) to the right. Make sure your lines extend indefinitely.
5Step 5: Check Key Points
Ensure the accuracy of your graph by checking a few key points. For example, at \(x = 1\), calculate \(y = -|1| - 2 = -3\), so the point (1,-3) should be on the graph. Similarly, at \(x = -1\), \(y = -|-1| - 2 = -3\), so the point (-1,-3) should also lie on the graph.
Key Concepts
Vertex of an Absolute Value FunctionTransformations of FunctionsSymmetry in Graphs
Vertex of an Absolute Value Function
The vertex of an absolute value function is the point where the graph changes direction. In the function \(y = |x|\), the vertex is at the origin, which means the graph has a turning point at (0, 0). For the function \(y = -|x| - 2\), transformations are applied to move the vertex.
- The negative sign in front affects the orientation (up or down) of the graph.
- Constant terms, like \(-2\), shift the vertex vertically.
Transformations of Functions
Transformations are changes made to the basic graph of a function. For absolute value functions, these transformations can include reflections, translations, and stretches or compressions. In our exercise function, \(y = -|x| - 2\), two primary transformations occur.
- Reflection over the x-axis: The absolute value function \(y = |x|\) normally opens upwards. However, the negative sign outside the absolute value causes a reflection over the x-axis, making the graph open downwards instead.
- Vertical shift: The \-2\ represents a vertical downward shift. This moves the entire graph down by 2 units on the y-axis, including the vertex from (0,0) to (0,-2).
Symmetry in Graphs
Symmetry is an important concept in graphing absolute value functions. A graph is symmetric if one half is a mirror image of the other. The absolute value function \(y = |x|\) is symmetric with respect to the y-axis. This means whatever is on one side of the y-axis is reflected or mirrored on the other side.
In our transformed function \(y = -|x| - 2\), this symmetry still exists, even after transformations. When the function is reflected and shifted, the graph remains symmetric about the y-axis.
In our transformed function \(y = -|x| - 2\), this symmetry still exists, even after transformations. When the function is reflected and shifted, the graph remains symmetric about the y-axis.
- The left side of the graph mirrors the right side.
- The highest point, or vertex, (0,-2) is still directly on the y-axis.
Other exercises in this chapter
Problem 21
For the following exercises, use a graphing utility to determine whether each function is one-to-one. $$ f(x)=-5 x+1 $$
View solution Problem 21
For the following exercises, determine the interval \((s)\) on which the function is increasing and decreasing. $$ g(x)=5(x+3)^{2}-2 $$
View solution Problem 21
Given \(f(x)=\sqrt{2-4 x}\) and \(g(x)=-\frac{3}{x},\) find the following: a. \((g \circ f)(x)\) b. the domain of \((g \circ f)(x)\) in interval notation
View solution Problem 21
Determine the interval(s) on which the function is increasing and decreasing. $$g(x)=5(x+3)^{2}-2$$
View solution